isomorphic spectral sequences => quasi-isomorphic filtered chain complexes?
Say your two filtered chain complexes are concentrated in degree zero. Then the spectral sequences degerate, and your questions become: If you have two filtered abelian groups and an isomorphism between the associated graded modules, can you deduce that the abelian groups are isomorphic? The answer is no; you can take $$ 0 \subset 2 \mathbb{Z} \subset \mathbb{Z} $$ and $$ 0 \subset \mathbb{Z} \subset \mathbb{Z} \oplus \mathbb{Z}/2. $$ As Ralph says, you usually need something on the chain level, unless you're in very degenerate cases (where the spectral sequence determines the isomorphism type of the object in the derived category).
What can be said is that if $C,C'$ have degree-wise finite filtrations and if $f: C \to C'$ is a filtration preserving chain map such that there is $r$ such that $E^r(f): E^r_{ij}(C) \to E^r_{ij}(C')$ is an isomorphism for all $i,j$ then $f$ is a quasi-isomorphm. This can be found in Brown: Cohomology of Groups, VII, Prop. 2.6.
Special cases where $E^\infty(C) \cong E^\infty(C')$ implies $H_\ast(C) \cong H_\ast(C')$ include:
The spectral sequence degenerates, e.g. $E^\infty_{ij}=0$ if $j \neq 0$ because in this case $H^i(C)=E^\infty_{i,0}$.
If $C,C'$ are complexes of modules over a ring $R$ and $E^\infty(C)$ consists of projective $R$-modules. In this case there is no extension problem and we have $H^n(C)=\oplus_{i+j=n}E^\infty_{ij}(C)$.
Here it is essential to consider the $E^\infty$-term since it can happen that $E^r(C) \cong E^r(C')$ but $E^{r+1}(C) \not\cong E^{r+1}(C')$. As an example take the LHS spectral sequence of the group extension
$$0 \to \mathbb{Z}/2 \to G \to \mathbb{Z}/2 \oplus \mathbb{Z}/2 \to 0.$$
Both, the dihedral group $D_8$ and the quaternion group $Q_8$ fit into this extension. Hence the spectral sequences have the same $E_2$-term but their $E_3$-terms differ.